Luo Gu ---- P1162 color Tiantu

https://www.luogu.org/problem/P1162

Digital 0 matrix consisting of 0, an arbitrary shape with a closed loop, a digital closed loop 1 is constituted only when the enclosure onto the left and right lower . 4 in four directions. Now requires all the spaces are filled into the closed circle 2 2 For example: 6 \ 6 Times 6 × matrix 6 ( n-6 = n- = 6), and the square before coloring the coloring follows:

0 0 0 0 0 0
0 0 1 1 1 1
0 1 1 0 0 1
1 1 0 0 0 1
1 0 0 0 0 1
1 1 1 1 1 1

0 0 0 0 0 0
0 0 1 1 1 1
0 1 1 2 2 1
1 1 2 2 2 1
1 2 2 2 2 1
1 1 1 1 1 1

Input Format

Each test line of an integer n-(. 1 \ n-Le \ Le 30) n- ( . 1 ≤ n- ≤ . 3 0 )

Next, n- n-row by 0 0 and . 1 1s of n-\ Times n- n- × n-square matrix.

Only a closed circle within a square, a circle at least 0 0

// thank U drinks Huang pointed out that this question is not the same data and data formats. Modified (input format)

Output Format

We have filled digital 2 complete phalanx 2.

Sample input and output

Input # 1copy

6
0 0 0 0 0 0
0 0 1 1 1 1
0 1 1 0 0 1
1 1 0 0 0 1
1 0 0 0 0 1
1 1 1 1 1 1

Output # 1copy

0 0 0 0 0 0
0 0 1 1 1 1
0 1 1 2 2 1
1 1 2 2 2 1
1 2 2 2 2 1
1 1 1 1 1 1

Description / Tips

1 \ n \ le 30 1 ≤ n ≤ 3 0

Solution: seeking direct circle 0 poor demand, we can mark out of the loop 0, 0 mark coming out of the loop completely, the rest is circle 0;

Starting from 0 for each vertical and horizontal boundary

AC Code:

#include<bits/stdc++.h>
using namespace std;
int arr[33][33];
struct stu{
    int x,y;
};
int n;
int d[4][2]={1,0,0,1,-1,0,0,-1};
void bfs(int x,int y){
    queue<stu >que;
    arr[x][y]=3;
    que.push({x,y});
    while(que.size()){
        stu s=que.front();
        que.pop();
        for(int i=0;i<4;i++){
            int dx=s.x+d[i][0];
            int dy=s.y+d[i][1];
            if(dx>=1&&dy>=1&&dx<=n&&dy<=n&&arr[dx][dy]==0){
                arr[dx][dy]=3;
                que.push({dx,dy});
            }
        }
    }
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%d",&arr[i][j]);
    for(int i=1;i<=n;i++){
        if(arr[i][1]==0) bfs(i,1);
        if(arr[i][n]==0) bfs(i,n);
        if(arr[1][i]==0) bfs(1,i);
        if(arr[n][i]==0) bfs(n,i);
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++)
            if(arr[i][j]==3) cout<<0<<" ";
            else if(arr[i][j]==0) cout<<2<<" ";
            else cout<<1<<" "; 
        cout<<endl;
    }
    return 0;
}

 I want the same type of subject: https://www.luogu.org/problem/P1506